Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int arr[150][150];
int n, m;
void sub_row(int row, int val) {
  for (int i = 0; i < m; ++i) {
    arr[row][i] -= val;
  }
}
void sub_col(int col, int val) {
  for (int i = 0; i < n; ++i) {
    arr[i][col] -= val;
  }
}
pair<int, int> max_col(int col) {
  int mn = INT_MAX;
  int mx = INT_MIN;
  for (int i = 0; i < n; ++i) {
    if (arr[i][col] < mn) {
      mn = arr[i][col];
    }
    if (arr[i][col] > mx) {
      mx = arr[i][col];
    }
  }
  pair<int, int> tmp = {mn, mx};
  return tmp;
}
pair<int, int> max_row(int row) {
  int mn = INT_MAX;
  int mx = INT_MIN;
  for (int i = 0; i < m; ++i) {
    if (arr[row][i] < mn) {
      mn = arr[row][i];
    }
    if (arr[row][i] > mx) {
      mx = arr[row][i];
    }
  }
  pair<int, int> tmp = {mn, mx};
  return tmp;
}
int main() {
  cin >> n >> m;
  vector<int> mn_row(n);
  for (int i = 0; i < n; ++i) {
    int mn = INT_MAX;
    for (int j = 0; j < m; ++j) {
      cin >> arr[i][j];
      if (arr[i][j] < mn) {
        mn = arr[i][j];
      }
    }
    mn_row[i] = mn;
  }
  vector<string> res;
  if (n <= m) {
    for (int i = 0; i < n; ++i) {
      if (mn_row[i] > 0) {
        sub_row(i, mn_row[i]);
        for (int j = 0; j < mn_row[i]; ++j) {
          res.push_back("row " + to_string(i + 1));
        }
      }
    }
    for (int i = 0; i < m; ++i) {
      pair<int, int> temp = max_col(i);
      if (temp.first != temp.second) {
        cout << -1 << "\n";
        return 0;
      }
      for (int j = 0; j < temp.first; ++j) {
        res.push_back("col " + to_string(i + 1));
      }
    }
  } else {
    for (int i = 0; i < m; ++i) {
      pair<int, int> val = max_col(i);
      if (val.first > 0) {
        sub_col(i, val.first);
        for (int j = 0; j < val.first; ++j) {
          res.push_back("col " + to_string(i + 1));
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      pair<int, int> val = max_row(i);
      if (val.first != val.second) {
        cout << -1 << "\n";
        return 0;
      }
      for (int j = 0; j < val.first; ++j) {
        res.push_back("row " + to_string(i + 1));
      }
    }
  }
  cout << res.size() << "\n";
  for (string str : res) {
    cout << str << "\n";
  }
  return 0;
}